An object with a mass of $125 g$ is dropped into $720 mL$ of water at $0^@C$ . If the object cools by $40 ^@C$ and the water warms by $6 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-05-24T05:26:53

The specific heat is $=3.62kJkg^-1K^-1$

The heat is transferred from the hot object to the cold water.

For the cold water, $Delta T_w=6ºC$

For the object $DeltaT_o=40ºC$

$m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)$

$C_w=4.186kJkg^-1K^-1$

Let, $C_o$ be the specific heat of the object

$0.125*C_o*40=0.72*4.186*6$

$C_o=(0.72*4.186*6)/(0.125*40)$

$=3.62kJkg^-1K^-1$

An object with a mass of 125 g125 g is dropped into 720 mL720 mL of water at 0^@C0^@C. If the object cools by 40 ^@C40 ^@C and the water warms by 6 ^@C6 ^@C, what is the specific heat of the material that the object is made of?